Testwiki:Reference desk/Archives/Mathematics/2012 October 15

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October 15

Indefinite integral of inverse function

If we know the indefinite integral g(x) = $\int f(x)dx,$ can we find the indefinite integral $\int f^{-1}(y)dy$ directly from g(x) without performing an integration? Duoduoduo (talk) 23:20, 15 October 2012 (UTC)

Integration by parts shows that

$\underset{f(a)}{\overset{f(x)}{\int }}{f}^{-1}(t)dt+\underset{a}{\overset{x}{\int }}f(s)ds=xf(x)-af(a).$ From this you should be able to figure out whatever you need to.Sławomir Biały (talk) 00:10, 16 October 2012 (UTC)

So $\int f^{-1}(x)dx=x{f}^{-1}(x)-g(f(x))+C$; example: $\int \arcsin xdx=x\arcsin x+\cos (\arcsin x)+C=x\arcsin x+\sqrt{1-{\sin }^2(\arcsin x)}+C=x\arcsin x+\sqrt{1-x^2}+C$ Ssscienccce (talk) 06:33, 17 October 2012 (UTC)

Thanks. Am I correct in assuming there's a typo in your first equation, Ssscienccce -- f(x) should be f ^-1(x)? Duoduoduo (talk) 20:56, 17 October 2012 (UTC)